<p>Let \(f:(0,\infty) \to R\) be a differentiable function satisfying \(f(x) + e^{f(x)} = \dfrac{2}{x} - \ln x - 1\). Find the number of integers in the range of \(x\) satisfying the inequality \(f(2x^2+1) - f(x^2+5) \geq f(1),\; x > 0\).</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) + e^f(x) = 2/x - ln(x) - 1 defines f implicitly; differentiating both sides reveals f'(x) = -1/[x(1+e^f(x))], which is always negative, making f strictly decreasing. Use this monotonicity to convert the inequality into a simpler algebraic condition.
<p><strong>Step 1: Establish monotonicity via differentiation</strong></p><p>Differentiate both sides of f(x) + e^f(x) = 2/x - ln(x) - 1:</p><p>f'(x) + e^f(x)·f'(x) = -2/x² - 1/x</p><p>f'(x)[1 + e^f(x)] = -(2/x² + 1/x) = -(2 + x)/(x²)</p><p>Therefore: f'(x) = -(2+x)/[x²(1+e^f(x))]</p><p><strong>Step 2: Analyze the sign of f'(x)</strong></p><p>Since x > 0, we have: 2+x > 0, x² > 0, and e^f(x) > 0, so 1+e^f(x) > 1 > 0</p><p>Thus f'(x) < 0 for all x > 0. This means <strong>f is strictly decreasing</strong> on (0,∞)</p><p><strong>Step 3: Convert inequality using monotonicity</strong></p><p>f(2x²+1) - f(x²+5) ≥ f(1)</p><p>f(2x²+1) ≥ f(x²+5) + f(1)</p><p>Since f is strictly decreasing, we need to establish a relationship. Note that at x=1: f(1) + e^f(1) = 2 - 0 - 1 = 1</p><p>Rearrange the original inequality: f(2x²+1) - f(x²+5) ≥ f(1)</p><p>Since f is decreasing: f(2x²+1) ≥ f(x²+5) requires 2x²+1 ≤ x²+5</p><p><strong>Step 4: Solve the algebraic inequality</strong></p><p>2x² + 1 ≤ x² + 5</p><p>x² ≤ 4</p><p>x² ≤ 4 (with x > 0)</p><p>0 < x ≤ 2</p><p><strong>Step 5: Count integers in the range</strong></p><p>For 0 < x ≤ 2, the integers are x ∈ {1, 2}</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2